NCERT Exemplar MCQs

Hint

(c)  The concept of circular paths of fixed energy was proposed by Bohr and not derived from Rutherford’s scattering experiment.

Hint

\(
\text { (b) Correct configuration in ground state should be } 1 s^2 2 s^2 2 p^6 3 s^2 3 p^{10} 3 d^{10} 4 s^1
\)

Hint: Fully filled subshell is more stable than partially filled subshell
Correct configuration should be \(1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^{10} 4 s^1\) for the copper which has atomic number 29 \(
\left({ }_{29} \mathrm{Cu}\right)\). Due to the extra stability of fully filled orbital of \(\)d-\(\) subshell, the last electron enters into the d-orbital instead of the s-orbital. Hence, option second is the correct answer.

Hint

Hint: The probability of finding an electron decreases uniformly as the distance from the nucleus increases
The probability density of electrons in \(2 \mathrm{~s}\) orbital first increases then decreases and after that, it begins to increase again as distance increases from the nucleus.

Hint

(d) Cathode rays consist of negatively charged material particles called electrons. They were discovered by William Crookes. The characteristics of cathode rays do not depend upon the material of electrodes and the nature of the gas present in the cathode ray tube.

Hint

The mass of the electron is very small as compared to the mass of the neutron.
Mass of electron \(=9.1 \times 10^{-31} \mathrm{~kg}\)
Mass of neutron \(=1.67 \times 10^{-27} \mathrm{~kg}\)

Hint

Hint: Atom is neutral
J J Thomson, in 1898, proposed the plum pudding, (raisin pudding or watermelon) model of an atom. An important feature of this model is that the mass of the atom is assumed to be uniformly distributed over the atom. This model was able to explain the overall neutrality of the atom.

Hint

Hint: Elements have the same mass number and different atomic number Isobars have the same mass number (i.e., the sum of protons and neutrons) but a different atomic number (i.e., number of protons). e.g., \({ }_{18}^{40} \mathrm{Ar}\) and \({ }_{19}^{40} \mathrm{~K}\) are isobars.
\(\begin{array}{ll}{ }_{18}^{40} \mathrm{Ar} & { }_{19}^{40} \mathrm{~K}\end{array}\)
Atomic number \(=18\) Atomic number \(=19\)
Mass number \(=40 \quad\) Mass number \(=40\)

Hint

Explanation:
Quantum numbers are characteristic quantities that are used to describe the various properties of an electron in an atom-like position, energy, or spin of the electrons. There are four quantum numbers namely, principal quantum number, azimuthal quantum number, magnetic quantum number, and spin quantum number.
Now for an orbital, the total number of nodes is \((n-1)\). Where \(\mathrm{n}\) is the principal quantum number. The principal quantum numbers of the shells are, for \(\mathrm{K}\) shell is 1 , for \(\mathrm{L}\) is 2 , for \(\mathrm{M}\) shell is 3 , for \(\mathrm{N}\) shell is 4 , etc. it cannot be fraction number.
Now the number of angular nodes is equal to the azimuthal quantum number denoted by ‘ \(\mathrm{l}\) ‘.
Therefore, the total number of radial nodes is, \((n-l-1)\)
Now, for \(3 p\) orbital, the value of \(\mathrm{n}\) is 3 , and the azimuthal quantum number is 1 . Therefore, the number of radial node is,
\(
\begin{aligned}
& (n-l-1) \\
= & (3-1-1) =1
\end{aligned}
\)

Hint

Hint: Angular node is equal to I value
Number of angular nodes \(=1\)
For \(4^{\text {th }}\) orbital \((n=4)\) and \(l=2\) for \(d\)-orbital
Number of angular nodes \(=2\)

Explanation: A radial node is a sphere that occurs when the radial wave function of the atomic orbital is zero or the sign of the wave function changes. On the other hand, angular nodes are either \(x, y\) or \(z\) planes where the electrons aren’t present.
The number of radial nodes can be solved on the basis of the following equation:
\(
{r}={n}-{l}-1
\)
The formula for the total number of nodes is, \({n}-1\)
Therefore angular nodes \(=\) total nodes-radial nodes.
For the \(4 \mathrm{~d}\) orbital, \({n}=4\) and \({l}=2\)
Therefore, \(r=4-2-1; r=1\).
Total number of nodes \(=4-1=3\)
Therefore the angular nodes \(=\) total nodes – radial nodes \(=3-1=2\)
Therefore the \(4 \mathrm{~d}\) orbital has 1 radial node and 2 angular nodes.

Hint

(b) According to Heisenberg’s uncertainty principle, the position and velocity of an electron cannot be determined simultaneously with accuracy which rules out the existence of fixed paths.

Explanation: Hint: Position of electron and momentum of electron cannot calculate simultaneously
Werner Heisenberg, a German physicist in 1927 , stated uncertainty principle which states that it is impossible to determine simultaneously, the exact position and exact momentum of an electron.
Mathematically, \(\Delta \mathrm{x} \times \Delta \mathrm{p} \geq \frac{\mathrm{h}}{4 \pi}\)
The important implications of the Heisenberg uncertainty principle is that it rules out existence of definite paths or trajectories of electrons and other similar particles.

Hint

\(
\text { (c) No of orbitals in } 3^{\text {rd }} \text { shell }(n=3)=n^2=3^2=9 \text {. }
\)

Explanation:

Hint: In atomic theory and quantum mechanics, an atomic orbital is a mathematical function describing the location and wave-like behaviour of an electron in an atom. This function can be used to calculate the probability of finding any electron of an atom in any specific region around the atom’s nucleus.

We have been asked about the number of orbitals associated with third shell, So, for that,
We know that there can be two electrons in one orbital maximum. The s sublevel has just one orbital, so can contain 2 electrons max. The p sublevel has 3 orbitals, so can contain 6 electrons max. The d sublevel has 5 orbitals, so can contain 10 electrons max.
Now, the orbit around the nucleus within which the electron rotates is called shells or Energy levels.” Each discrete distance orbit from the nucleus corresponds to a certain energy level. The electron that rotates in the lowest orbit has the lowest energy level and in the outermost orbit, electrons have higher energy levels.
We know that, the total number of orbitals in the nth shell: \(n^2\),
Where, \(n=\) shell/ principal quantum number,
So, we need to calculate the total number of orbitals in the third shell,
So, \(\mathrm{n}\) will be equal to 3 ,
Therefore, the orbitals would be: \(3^2=9\).
So, the orbitals are one 3-s orbital, three 3-p orbital and five 3-d orbitals.

Hint

\(
\text { (a) Orbital angular momentum }=\sqrt{l(l+1)} \frac{h}{2 \pi} \text {. Hence, it depends only on ‘ } l \text { ‘. }
\)
Example: What is the Angular Quantum Number of d electrons?
Ans: The orbital angular momentum is \((\mathrm{L})=\sqrt{l(l+1)} \frac{h}{2 \pi}\)
Here, the orbital \(d\) is used. So, the value of \(l=2\)
Therefore, the orbital angular momentum can be represented as \(\sqrt{6} \frac{h}{2 \pi}\)

Hint

\(
\text { Hint: average atomic mass }=\frac{(\mathrm{Cl}-37 \text { isotope } \% \times \text { atomic mass })+(\mathrm{Cl}-35 \text { isotope } \% \times \text { atomic mass })}{100}
\)
Step 1:
The fractional mass of chlorine is 35.5. The isotopic percentage of \(\mathrm{Cl}-37\) is \(\mathrm{X} \%\) and the isotopic percentage of \(\mathrm{Cl}-35\) is \(100-\mathrm{x} \%\). average atomic mass \(=\frac{(\mathrm{Cl}-37 \text { isotope } \% \times \text { atomic mass })+(\mathrm{Cl}-35 \text { isotope } \% \times \text { atomic mass })}{100}\)
Step 2:
Calculate the ratio is as follows:
\(
\begin{aligned}
& 35.5=\frac{(\mathrm{X} \times 37)+((100-\mathrm{X}) \times 35)}{100} \\
& 3550=37 \mathrm{X}+3500-35 \mathrm{X} \\
& 3550-3500=2 \mathrm{X} \\
& 50=2 \mathrm{X} \\
& \mathrm{X}=25 \\
& 100-\mathrm{X}=75
\end{aligned}
\)
The ratio of \(\mathrm{Cl}-37\) and \(\mathrm{Cl}-35=\frac{25}{75}\) \(=1: 3\)

Alternate:

\(
{ }_{17}^{37} \mathrm{Cl}:{ }_{17}^{35} \mathrm{Cl}
\)
\(
1: \quad \quad 3 \text { Ratio }
\)
\(
\begin{aligned}
\begin{array}{c}
\text { Average atomic mass }
\end{array} & =\frac{(37 \times 1)+(35 \times 3)}{1+3} \\
& =\frac{142}{4}=35.5
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \text { (b) } \mathrm{Cr}^{3+}(Z=24)=[\mathrm{Ar}] 3 d^3 \\
& \mathrm{Fe}^{3+}(Z=26)=[\mathrm{Ar}] 3 d^5 \\
& \mathrm{Mn}^{2+}(Z=25)=[\mathrm{Ar}] 3 d^5 \\
& \mathrm{Co}^{3-}(Z=27)=[\mathrm{Ar}] 3 d^6
\end{aligned}
\)
\(
\mathrm{Sc}^{3+}(Z=21)=[\mathrm{Ar}] 3 d^0
\)
\(\mathrm{Fe}^{3+}\) and \(\mathrm{Mn}^{2+}\) will have the same number of electrons, i.e., 23 and, hence, have the same electronic configuration.

Explanation:
Electronic configuration of \(\mathrm{Fe}^{3+}\)
– Electronic configuration of \(\mathrm{Fe}\) is \(1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 4 s^2 3 d^6\).
– Three electrons must be released for \(\mathrm{Fe}^{3+}\).
– So, the electronic configuration of \(\mathrm{Fe}^{3+}\) is \(1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^5\).
Electronic configuration of \(\mathrm{Mn}^{2+}\)
– Electronic configuration of \(\mathrm{Mn}\) is \(1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 4 s^2 3 d^5\).
– Two electrons must be released for \(\mathrm{Mn}^{2+}\).
– So, the electronic configuration of \(\mathrm{Mn}^{2+}\) is \(1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^5\).
So, pair of ions having the same electronic configuration is \(\mathrm{Fe}^{3+}, \mathrm{Mn}^{2+}\). The explanation for the incorrect options
– \(\mathrm{Cr}^{3+}, \mathrm{Fe}^{3+}\) :-The electronic configuration of \(\mathrm{Cr}^{3+}\) is \(1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^3\).
– The electronic configuration of \(\mathrm{Fe}^{3+}\) is \(1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^5\).
– \(\mathrm{Fe}^{3+}, \mathrm{Co}^{3+}\) :-The electronic configuration of \(\mathrm{Co}^{3+}\) is \(1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^6\).
– \(\mathrm{Sc}^{3+}, \mathrm{Cr}^{3+}\) : The electronic configuration of \(\mathrm{Sc}^{3+}\) is \(1 s^2 2 s^2 2 p^6 3 s^2 3 p^6\).
– Hence. options (a), (c), and (d) are incorrect.
Hence, option(b) is correct. The pair of ions having the same electronic configuration is \(\mathrm{Fe}^{3+}, \mathrm{Mn}^{2+}\).

Hint

\(
\text { Hint: Two electrons present in any orbital have spin quantum numbers } m_s \text { but of opposite sign, that is, } \frac{-1}{2} \text { and } \frac{1}{2} \text {. }
\)

(d) (a) Electrons in \(2 s\) and \(2 p\) orbitals have different screening effect. Hence, their \(Z_{\text {eff }}\) is different. \(Z_{\text {eff }}\) of \(2 s\) orbital \(>Z_{\text {eff }}\) of \(2 p\) orbital Therefore, it is not correct.
(b) Energy of \(2 s\) orbital \(<\) energy of \(2 p\) orbital. Hence, it is not correct.
(c) \(Z_{\text {eff }}\) of \(1 s\) orbital \(\neq Z_{\text {eff }}\) of 2 s orbital Hence, it is incorrect.
(d) For the two electrons of \(2 s\) orbital, the value of \(m_{\mathrm{s}}\) is \(+\frac{1}{2}\) and \(-\frac{1}{2}\). Hence, it is correct.

Hint

(b) \(\lambda=\frac{h}{m v}\). For same value of \(v\), larger the value of mass \(m\), shorter is the wavelength, \(\lambda\). Here, \(\alpha\)-particles have the largest mass.
Since, alpha particles have the largest mass, they have the shortest wavelength.

Hint

(a) \({ }_6^{12} \mathrm{X}\) and \({ }_6^{13} \mathrm{Y}\) have same atomic number but different mass number.
(b) \({ }_{17}^{35} \mathrm{X}\) and \({ }_{17}^{37} \mathrm{Y}\) have same atomic number but different mass number. Both these pairs are isotopes to each other.
(c) \({ }_6^{14} \mathrm{X}\) and \({ }_7^{14} \mathrm{Y}\) have different atomic number but same mass number.
(d) \({ }_4^8 \mathrm{X}\) and \({ }_5^8 \mathrm{Y}\) have different atomic number but same mass number.
(c, d) Isotopes have the same atomic number but different mass number.

Hint

(a, d) Degenerate orbitals mean the orbitals of the same sub-shell of the same main shell, i.e., their \(\mathrm{n}\) and l values are the same. Other two pairs have different values of \(n\) and \(l\) hence, cannot be having the same energy. i.e., their \(\mathrm{n}\) and l value.
(a) (i) \(3 \mathrm{~d}_{\mathrm{xy}}\) (ii) \(3 \mathrm{~d}_{\mathrm{yz}}\)
(b) (i) \(3 \mathrm{p}_{\mathrm{x}}\) (ii) \(3 \mathrm{~d}_{\mathrm{xy}}\)
(c) (i) \(4 \mathrm{~s}\) (ii) \(3 \mathrm{~d}_{\mathrm{xy}}\)
(d) (i) \(4 \mathrm{~d}_{\mathrm{x}^2-\mathrm{y}^2}\) (ii) \(3 \mathrm{~d}_{\mathrm{x}^2-\mathrm{y}^2}\)
Thus, \(3 \mathrm{~d}_{\mathrm{xy}}\) and \(3 \mathrm{~d}_{\mathrm{yz}} ; 3 \mathrm{~d}_{\mathrm{x}^2-\mathrm{y}^2}\) and \(3 \mathrm{~d}_{\mathrm{x}^2-\mathrm{y}^2}\) represent pair of degenerate orbitals.

Hint

(b, c) If \(n=1, l \neq 1\). Hence, (i) is wrong.
If \(n=2, l=0,1\). For \(l=1, m=-1,0,+1\). Hence (ii) is correct.
If \(n=3,l=0,1,2\). For \(l=2, m=-2,-1,0,+1,+2\). Hence (iii) is correct.
If \(n=3, l \neq 4\). Hence, (iv) is wrong.

Hint

For,
\(
\begin{aligned}
& \mathrm{Na}^{+}=11-1=10 \mathrm{e}^{-} \\
& \mathrm{Mg}^{2+}=12-2=10 \mathrm{e}^{-}
\end{aligned}
\)
Thus, they have same number of electrons.
\(
\text { Fo, } \mathrm{Al}^{3+}=13-3=10 \mathrm{e}^{-}, \mathrm{O}^{-}=8+1=9 \mathrm{e}^{-}
\)
They do not have same number of electrons.
For, \(\mathrm{Na}^{+}=10 \mathrm{e}^{-}, \mathrm{O}^{2-}=8+2=10 \mathrm{e}^{-}\)
They have same number of electrons.
For, \(\mathrm{N}^{3-}=7+3=10 \mathrm{e}^{-}, \mathrm{Cl}^{-}=17+1=18 \mathrm{e}^{-}\)
They do not have same number of electrons.
Thus, \(\mathrm{Na}^{+}\)is isoelectronic with \(\mathrm{Mg}^{2+}\) and \(\mathrm{O}^{2-}\).

Hint

(a, d)
(a) Azimuthal quantum number l is also known as orbital angular momentum or subsidiary quantum number. It determines the three-dimensional shape of the orbital.
(b) The principal quantum number determines the size of the orbit.
(c) Magnetic quantum number determines the orientation of the electron cloud in a subshell.
(d) An electron spins around its own axis, much in a similar way as the earth spins around its own axis while revolving around the sun. In other words, an electron has, besides charge and mass, intrinsic spin angular quantum number.

Hint

(a) s-orbitals shield the electrons from the nucleus more effectively than p-orbitals which in turn shield more effectively than d-orbitals. Hence, the arrangement of subshells in the increasing order of effective nuclear charge is:
\(d<p<s\)

Hint

From the orbital diagram, it is seen that there are two unpaired electrons.

Hint

\({ }_{28} \mathrm{Ni}=1 \mathrm{~s}^2, 2 \mathrm{~s}^2, 2 \mathrm{p}^6, 3 \mathrm{~s}^2, 3 \mathrm{p}^6, 3 \mathrm{~d}^8, 4 \mathrm{~s}^2\); Nickel will lose 2 electrons from \(4 \mathrm{~s}\) (outermost shell) to form \(\mathrm{Ni}^{2+}\) ion.
Hence, \({ }_{26} \mathrm{Ni}^{2+}=1 \mathrm{~s}^2, 2 \mathrm{~s}^2, 2 \mathrm{p}^6, 3 \mathrm{~s}^2, 3 \mathrm{p}^6, 3 \mathrm{~d}^8\)

Hint

Degenerate orbitals are the orbitals of the same subshell of the same main shell. Hence, these are
\(
\left(3 d_{x y}, 3 d_{z^2}, 3 d_{y z}\right) \text { and }\left(4 d_{x y}, 4 d_{x z}, 4 d_{z^2}\right) \text {. }
\)

Hint

For \(3 p\) orbital \(n=3, l=1\)
Number of angular nodes \(=l=1\)
Number of radial nodes \(=n-l-1=3-1-1=1\)

Hint

Atomic mass number \(=A=13, n=7\)
As \(A=n+p, p=A-n=13-7=6\)
Hence \(Z=p=6\)

Hint

Hint: \(\mathrm{Cu}(\mathrm{Z}=29): 1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^6 3 \mathrm{~s}^2 3 \mathrm{p}^6 3 \mathrm{~d}^{10} 4 \mathrm{~s}^1\)
The electronic configuration of the given atom or ions are as follows:
A. \(\mathrm{Cu}(\mathrm{Z}=29): 1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^6 3 \mathrm{~s}^2 3 \mathrm{p}^6 3 \mathrm{~d}^{10} 4 \mathrm{~s}^1\)
B. \(\mathrm{Cu}^{2+}(\mathrm{Z}=29): 1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^6 3 \mathrm{~s}^2 3 \mathrm{p}^6 3 \mathrm{~d}^9\)
C. \(\mathrm{Zn}^{2+}(\mathrm{Z}=30): 1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^6 3 \mathrm{~s}^2 3 \mathrm{p}^6 3 \mathrm{~d}^{10}\)
D. \(\mathrm{Cr}^{3+}(\mathrm{Z}=24): 1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^6 3 \mathrm{~s}^2 3 \mathrm{p}^6 3 \mathrm{~d}^3\)

Hint

Hint: Orientation of the orbital is represented by m values.
Explanation:
(i). Principal quantum number is the most important quantum number as it determines the size and to large extent the energy of the orbital.
(ii). Azimuthal quantum number determines the angular momentum of the electron and defines the three-dimensional shape of the orbital.
(iii). Magnetic quantum number gives information about the spatial orientation of orbitals with respect to a standard set of coordinate axes.
(iv). Spin quantum number arises from the spectral evidence that an electron in its motion around the nucleus in an orbit also rotates or spin about its own axis.

Hint

Hint: Heisenberg’s Uncertainty Principle states that it is impossible to determine the exact position and exact momentum of a subatomic particle simultaneously.
Explanation:
(i) Hund’s rule states that pairing of electrons in the orbitals belonging to the same subshell ( \(p\), \(d\) or \(f\) ) does not take place until each orbital belonging to that subshell has one electron each i.e., it is singly occupied.
(ii) Aufbau principle states that in the ground state of the atoms, the orbitals are filled in order of their increasing energies.
(iii) According to the Pauli exclusion principle, no two electrons in an atom can have the same set of four quantum numbers.
(iv) Heisenberg’s uncertainty principle states that it is impossible to determine the exact position and exact momentum of a subatomic particle simultaneously.

Hint

\(
\begin{array}{lll}
\text { Name } & \text { Frequency } & \text { Uses } \\
\text { (i) X-rays } & 2 \times 10^{16}-3 \times 10^{19} & \text { Medical pictures, medical } \\
& \mathrm{Hz} & \text { testing }
\end{array}
\)
\(
\text { (ii) Ultraviolet wave (UV) } 7.9 \times 10^{14}-2 \times 10^{16} \text { Germicidal lamp }
\)
\(
\begin{array}{lll}
\text { (iii) Long radio waves } & 10^0-10^4 \mathrm{~Hz} & \text { Signal transmission } \\
\text { (iv) Microwave } & 1 \times 10^9-5 \times 10^{11} & \text { Cooking radar }
\end{array}
\)

Hint

Hint: \(\psi^2\) tells about probability density
(i) Photon has particle nature as well as wave nature. It exhibits both momentum and wavelength.
(ii) Electron also has particle nature as well as wave nature. Thus, it also exhibits both momentum and wavelength.
(iii) \(\psi^2\) represents the probability density of electrons and always has positive values.
(iv) Principle quantum number \(\mathrm{n}=4\) for \(\mathrm{N}\)-shell.
K L M N
\(\mathrm{n}=1234\)
It always has positive values.

Hint

Hint: Electronic configuration of \(\mathrm{Cr}\) is \(1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^6 3 \mathrm{~s}^2 3 \mathrm{p}^6 3 \mathrm{~d}^5 4 \mathrm{~s}^1\)
(i) \(\mathrm{Cr}(\mathrm{Z}=24)=1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^6 3 \mathrm{~s}^2 3 \mathrm{p}^6 3 \mathrm{~d}^5 4 \mathrm{~s}^1=[\mathrm{Ar}] 3 \mathrm{~d}^5 4 \mathrm{~s}^1\)
(ii) \(\mathrm{Fe}^{2+}(\mathrm{Z}=26)=1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^6 3 \mathrm{~s}^2 3 \mathrm{p}^6 3 \mathrm{~d}^6 4 \mathrm{~s}^0=[\mathrm{Ar}] 3 \mathrm{~d}^6 4 \mathrm{~s}^0\)
(iii) \(\mathrm{Ni}^{2+}(\mathrm{Z}=28)=1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^6 3 \mathrm{~s}^2 3 \mathrm{p}^6 3 \mathrm{~d}^8 4 \mathrm{~s}^0=[\mathrm{Ar}] 3 \mathrm{~d}^8 4 \mathrm{~s}^0\)
(iv) \(\mathrm{Cu}(\mathrm{Z}=29)=1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^6 3 \mathrm{~s}^2 3 \mathrm{p}^6 3 \mathrm{~d}^{10} 4 \mathrm{~s}^1=[\mathrm{Ar}] 3 \mathrm{~d}^{10} 4 \mathrm{~s}^1\)

Hint

(a) Hint Chemical property depends on the number of electrons
Atoms with the same atomic number but with different atomic masses are called isotopes. In isotopes, a number of electrons are the same and these electrons are responsible for their chemical behaviour.
Hence, isotopes exhibit similar chemical properties

Hint

(b) Hint: With the temperature intensity of light increases.
A body that absorbs and emits all radiations falling on it is called a perfect black body. With rise in temperature, the frequency of radiation emitted by the body increases.

Hint

(iii) Hint: Heisenberg’s uncertainty principle.
Explanation:
According to Heisenberg’s uncertainty principle, the exact position and exact momentum of an electron cannot be determined simultaneously. The position of the electron and its path is not defined clearly in an atom.
Hence, the Assertion is true but the reason is false.